About the proof using a simple epsilon delta.

Question

Explanation

I want to prove this equation (1) using the $\epsilon-\delta$ definition for the function $f(x) = x^2$.

$$ \lim_{x \to a} f(x) = a^2 \tag{1} $$

I've read certain books and I don't agree with some of them.
In that book, $\delta$ was assumed as follows.

$$ \delta = \frac{\epsilon}{(1+2|a|+\epsilon)} $$

And the continuity can be proved by the following equation.

$$ \begin{aligned} |f(x) - f(a)| &= |x+a||x-a| \\ &< (|x|+|a|)\delta \\ &< (|a|+\delta + |a|)\delta \\ &= (\delta + 2|a|)\delta < \epsilon \end{aligned} \tag{2} $$

Questions

01

The first question is, why can this equation (2) prove that continuity holds?
There is no specific delta value for epsilon.

02

The first question is, why does equation (3) below hold?

$$ (\delta + 2|a|)\delta < \epsilon \tag{3} $$

Answer 1

The first expression from the book of $ \delta $ doesn't seem to be correct.

The inequality $ (2)$ does not prove continuity since it does not give the value of $ \delta $.

The inequality $ (3) $ comes from $$|x-a|<\delta \implies$$ $$-\delta<x-a<\delta \implies$$

$$-|a|-\delta\le a-\delta<x<a+\delta<|a|+\delta \implies$$

$$|x|<|a|+\delta\;\implies$$ $$|x|+|a|<2|a|+\delta \;\implies$$ $$|x^2-a^2|=|x+a||x-a|\le (|x|+|a|)|x-a|$$

$$\le (\delta+2|a|)|x-a|<(\delta+2|a|)\delta$$ So, to realise $|x^2-a^2|<\epsilon$, it is sufficient to have

$$(2|a|+\delta)\delta<\epsilon$$ but it does not give $\delta$.

You must add an other condition, like $|x-a|<\color{red}{1}$ .

With this addiitional condition, we will have

$$|x|+|a|<1+2|a|$$ and $$|x^2-a^2|<(1+2|a|)|x-a|$$ So, if you want that $$|x^2-a^2|<\epsilon$$ We will look for $ \delta$ such that

$$|x-a|<\delta\implies (1+2|a|)|x-a|<\epsilon$$ or $$|x-a|<\delta\implies |x-a|<\frac{\epsilon}{1+2|a|)}$$ thus, we can take $$\delta=\frac{\epsilon}{1+2|a|}$$ But, there is the additional condition $$|x-a|<\color{red}{1}$$ So, $\delta $ must be $\le 1$,

$$\delta=\min(\color{red}{1},\frac{\epsilon}{1+2|a|})$$ This choice, will prove continuity.

Answer 2

For the first question, according to what you reported, the book does give a specific value of $\delta$ for each value of $\epsilon.$ That value is $$ \delta = \frac{\epsilon}{1+2|a|+\epsilon}. $$

Notice that $1+2|a|+\epsilon > \epsilon > 0$ and hence this formula for $\delta$ ensures that $0 < \delta < 1.$

For the second question, the book makes several claims for the case where $\epsilon > 0,$ $\delta = \frac{\epsilon}{1+2|a|+\epsilon},$ and $|x-a| < \delta.$ One claim is

$$ |f(x) - f(a)| = |x+a||x-a|. $$

This is true because $m^2 - n^2 = (m+n)(m-n).$ Another claim is

$$ |x+a||x-a| < (|x|+|a|)\delta. $$

This is true because $|x-a| < \delta.$ Another claim is

$$ (|x|+|a|)\delta < (|a|+\delta + |a|)\delta . $$

This is true because $|x| < |a|+\delta.$ Then there is the claim that $ (|a|+\delta + |a|)\delta = (\delta + 2|a|)\delta, $ which is true by simple rearrangement and combination of terms. Finally, the book claims that

$$ (\delta + 2|a|)\delta < \epsilon. $$

Plugging in the chosen value of $\delta,$ this is equivalent to the claim that

$$ \left(\frac{\epsilon}{1+2|a|+\epsilon} + 2|a|\right) \frac{\epsilon}{1+2|a|+\epsilon} < \epsilon. $$

Working this out,

$$ 0 < \frac{\epsilon}{1+2|a|+\epsilon} < 1 $$

and

$$ 0 < \frac{1 + 2|a|}{1+2|a|+\epsilon} < 1, $$

and therefore

\begin{align} \left(\frac{\epsilon}{1+2|a|+\epsilon} + 2|a|\right) \frac{\epsilon}{1+2|a|+\epsilon} &< \left(1 + 2|a|\right) \frac{\epsilon}{1+2|a|+\epsilon} \\ &= \frac{1 + 2|a|}{1+2|a|+\epsilon}\epsilon \\ &< \epsilon . \end{align}

So the claim is true, but what an excessively complicated way this is to do it.

You get a simpler proof if you choose $\delta$ according to the formula in another answer, $$\delta < \min\left(1, \frac{\epsilon}{1+2|a|}\right).$$

The value of $\delta$ you get for any $\epsilon$ in this way may be greater than the $\delta$ selected by the book, but it is still small enough to finish the proof. See this answer for another proof.